= Solution
Set
$$
k_0=\frac{\sqrt{2mE}}\hbar,\qquad k_1=\frac{\sqrt{2m(E-V_1)}}\hbar,\qquad k_2=\frac{\sqrt{2m(E-V_2)}}\hbar.
$$
All are real and $k_0>k_1>k_2>0$. Suppressing a common time <quantum phase> $e^{-iEt/\hbar}$, choose
$$
\boxed{\psi_{\mathrm L}=Ae^{ik_0x}+Be^{-ik_0x},\qquad \psi_{\mathrm M}=Ce^{ik_1x}+De^{-ik_1x},\qquad \psi_{\mathrm R}=Fe^{ik_2(x-a)}.}
$$
The rightmost region has no left-moving term because no wave is incident from the right. The shifted <quantum phase> convention makes $F$ the transmitted <amplitude> at the second interface.
For finite potential steps and equal <mass>, <wavefunction matching at a finite potential step> requires continuity of both $\psi$ and $\partial_x\psi$. Explicitly,
$$
A+B=C+D,\quad k_0(A-B)=k_1(C-D),
$$
$$
Ce^{ik_1a}+De^{-ik_1a}=F,\quad k_1(Ce^{ik_1a}-De^{-ik_1a})=k_2F.
$$
A discontinuous <wavefunction> would introduce an inadmissible singular kinetic term, while integrating the stationary <Schrödinger equation> across a finite step shows that the <derivative> cannot jump. These conditions also keep the <probability current> continuous, so a finite interface neither creates nor destroys <probability>. The <derivative> rule would need modification for a delta-function potential or a different effective <mass>; neither occurs here.
Back to article page
